Question

"Question 2 In the following figure, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC). Can you answer the question that you have left in the ’Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area? [Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide ΔABC into n triangles of equal areas.]

Class 9 - Math - Areas of Parallelograms and Triangles Page 164"

Answer 1

Given: ABC is a Triangle , D & E are two Points on BC, Such that BD= DE= EC

To Prove:

ar (ABD) = ar (ADE) = ar (AEC)

Proof:

Let AO be the perpendicular to BC.

We know that,

Area of ∆ =1/2× base × height

ar(∆ABD)= ½× BD× AO

ar(∆ADE)= ½× DE× AO

ar(∆AEC)= ½× EC× AO

BD= DE= EC [given]

ar(∆ABD)= ar(∆ADE)= ar(∆AEC)

 

Yes Budhia divided the land in three equal parts ,as she used the result of this question.

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Hope this will help you...

 

Answer 2

Hi friend ✨✨

Let AL be perpendicular to BC .So, AL is the height of triangle ABD,ADE and AEC .

ar (ABD)=1/2×BD×AL

ar(ADE)=1/2×DE×AL

ar(AEC)=1/2×EC×AL

Since,. BD=DE=EC

ar(ABD)=ar(ADE)=ar (AEC)

Yups, altitudes of all ∆ are same .Budhia has use the result of this question.

Hope it helps...